2012-01-05T09:08:46-05:00
The line L is tangent to a circle, center at (2.5), at point (-1,1). Suppose line L is passing through (-5,4). Determine the equation of the line L.
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2012-01-06T09:08:20-0500
The line L is tangent to a circle, center at (2.5), at point (-1,1). Suppose line L is passing through (-5,4). Determine the equation of the line L. If the required line L passes through the points (-1,1) and (-5,4), then it's equation is (x-(-5)) / (-1-(-5)) = (y-4) / (1-4) (x+5)/4 = (y-4)/(-3) -3(x+5) = 4(y-4) -3x - 15 = 4y - 16 4y + 3x = 1. Let's check whether this line is tangent to a circle center at (2,5). If so, it will be normal to the line that passes through (2,5) and (-1,1): (x-(-1)) / (2-(-1)) = (y-1) / (5-1) (x+1)/3 = (y-1)/4 4(x+1) = 3(y-1) 4x + 4 = 3y - 3 3y - 4x = 7. Here is the the scalar product of the direction vectors: (4,3)*(3,-4) = 4*3 + 3*(-4) = 0. We see that it is tangent to a circle. So, required line equation is 4y + 3x = 1.
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